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<math>= \frac{3(e^{j4\pi t} - e^{-j4\pi t})}{2j}+\frac{2(e^{j4\pi t} + e^{-j4\pi t})}{2}</math> | <math>= \frac{3(e^{j4\pi t} - e^{-j4\pi t})}{2j}+\frac{2(e^{j4\pi t} + e^{-j4\pi t})}{2}</math> | ||
| + | <font size="4.5"> | ||
<math>=3sin(4\pi t) + 2 cos(4\pi t)</math> | <math>=3sin(4\pi t) + 2 cos(4\pi t)</math> | ||
| + | </font> | ||
Revision as of 13:55, 6 October 2008
Inverse Fourier Transform
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega $
$ X(\omega) = \pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j) $
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{j\omega t}d\omega $
$ =\frac{2-3j}{2}\int_{-\infty}^{\infty}\delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}\delta(\omega + 4\pi)e^{j\omega t}d\omega $
$ =\frac{2-3j}{2}e^{j4\pi t} + \frac{2+3j}{2}e^{-j4\pi t} $
$ =e^{j4\pi t}-\frac{3j}{2}e^{j4\pi t} + e^{-j4\pi t}+\frac{3j}{2}e^{-j4\pi t} $
$ =\frac{3}{2j}e^{j4\pi t}-\frac{3}{2j}e^{-j4\pi t}+e^{j4\pi t} + e^{-j4\pi t} $
$ = \frac{3(e^{j4\pi t} - e^{-j4\pi t})}{2j}+\frac{2(e^{j4\pi t} + e^{-j4\pi t})}{2} $
$ =3sin(4\pi t) + 2 cos(4\pi t) $
